Example of G and H in Banach Space s.t. $(G\cap H)^{\perp}\neq \overline{G^{\perp}+H^{\perp}}$

Question

I wanted to know example of G and H are closed subspaces of Banach space such that $(G\cap H)^{\perp}\neq \overline{(G^{\perp}+H^{\perp})}$

Here $G{\perp} $ is annihilator of G.

$G^{\perp}=\{f\in E^*|f(x)=0\forall x\in G\}$

I know that $\overline{G^{\perp}+H^{\perp}}\subset G\cap H $ But i do not know example where strict inequality holds

I think the example where $Z^{\perp}\subset (Z^{\perp})^{\perp}$ will work But I did not know that

I do not know how to tackle such problem.If you also tell me how to find such example that would be really helpful for me.

Any Help will be appreciated

Answer 1

If $x \in G \cap H$ then $x \perp G^{\perp}$ and $x \perp H^{\perp}$ so $x \perp (G^{\perp}+H^{\perp})$. [ If $g \in G^{\perp}$ and $h \in H^{\perp}$ then $g$ vanishes at $x$ and so does $h$. Hence $(g+h)(x)=0$]. This proves that LHS $\subseteq$ RHS.

Answer for the revised question:

I will write $M^{\perp}$ for $(\{x \in E: f(x)=0 \forall f \in M\})$ for subspaces $M$ of the dual space. Then $(G^{\perp}+H^{\perp})^{\perp} \subseteq G \cap H$. Hence $(G \cap H)^{\perp} \subset \overline {G^{\perp}+H^{\perp}}$ which proves that the counterexample you are looking for does not exist.